A + 2B rightarrow AB_{2}. 36.0 g of 'A' (Mol | vedclass

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(A) $1$. Initial moles: $n_A = \frac{36.0 \ g}{60 \ g \ mol^{-1}} = 0.6 \ mol$, $n_B = \frac{56.0 \ g}{80 \ g \ mol^{-1}} = 0.7 \ mol$.$2$. Limiting r

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$B$, $C$ and $D$ only

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(A) $1$. Initial moles: $n_A = \frac{36.0 \ g}{60 \ g \ mol^{-1}} = 0.6 \ mol$, $n_B = \frac{56.0 \ g}{80 \ g \ mol^{-1}} = 0.7 \ mol$.$2$. Limiting reagent: For $A + 2B ightarrow AB_{2}$, $1 \ mol$ of $A$ requires $2 \ mol$ of $B$. $0.6 \ mol$ of $A$ requires $1.2 \ mol$ of $B$. Since we have only $0.7 \ mol$ of $B$, '$B$' is the limiting reagent. Statement $(A)$ is incorrect.$3$. Product formed: $0.7 \ mol$ of $B$ produces $\frac{0.7}{2} = 0.35 \ mol$ of $AB_{2}$. Molar mass of $AB_{2} = 60 + (2 	imes 80) = 220 \ g \ mol^{-1}$. Mass of $AB_{2} = 0.35 \ mol 	imes 220 \ g \ mol^{-1} = 77.0 \ g$. Statements $(B)$ and $(C)$ are correct.$4$. Unreacted '$A$': $0.7 \ mol$ of $B$ reacts with $\frac{0.7}{2} = 0.35 \ mol$ of $A$. Remaining $A = 0.6 - 0.35 = 0.25 \ mol$. Mass of remaining $A = 0.25 \ mol 	imes 60 \ g \ mol^{-1} = 15.0 \ g$. Statement $(D)$ is correct.Therefore, statements $(B)$, $(C)$, and $(D)$ are correct.

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